Rate and Work Problems

Learning goals

Turn a completion time into the rate $\tfrac{1}{T}$
Add rates, never times, when workers combine
Subtract known rates to find a missing worker's time
Treat a drain as a negative rate in the signed sum
Apply $d = rt$ , solving for whichever is unknown
Compute average speed from totals, not from a mean

From time to rate

Suppose that a job is done at a steady pace, and that it takes $T$ units of time from start to finish. Then in one unit of time exactly the fraction $\frac{1}{T}$ of that job gets done. That fraction per unit time is the work rate. A pump that empties a pool in $5$ hours removes one fifth of the pool each hour, so its rate is $\frac{1}{5}$ pool per hour. In general,

\text{rate} = \frac{1}{T} \quad \text{(job per unit time)}.

Rate and time are reciprocals, which is the inverse variation of the last lesson wearing new clothes. The amount of work is fixed at one whole job, so $\text{rate} \times \text{time} = 1$ , a constant product, which means the rate is inversely proportional to the time. Double the time and the rate halves; a faster rate goes with a shorter time. This reciprocal is the hinge of every work problem, so it is worth turning both ways:

\text{rate} = \frac{1}{\text{time}}, \qquad \text{time} = \frac{1}{\text{rate}}.

You almost always work in rates while solving, then flip the final rate back into a time at the very end.

Why work rates add

Rates combine cleanly in a way that times refuse to, and it is worth seeing exactly why before trusting it.

Why work rates add#

Picture two workers on the same job, one who would finish it alone in $T_1$ units of time and one who would finish it alone in $T_2$ units. Ask what happens in a single unit of time while both work at once. The first worker, going at the steady rate $\frac{1}{T_1}$ , completes $\frac{1}{T_1}$ of the job in that unit. The second, going at rate $\frac{1}{T_2}$ , completes $\frac{1}{T_2}$ of the job in the same unit. Assume the two do not get in each other’s way. Then whatever the first finishes is simply work the second no longer has to do, so the fractions they complete add together. In that one unit of time the team finishes

\frac{1}{T_1} + \frac{1}{T_2}

of the job. But the fraction of a job that a team finishes in one unit of time is, by the definition above, exactly the team’s combined rate. So the combined rate is the sum of the individual rates,

\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2},

where $T$ is the time the team takes together. Rates add; times do not. This is why you can never average the two times, nor add them: it is the rates, the work done per unit of time, that combine. Only once you have added the rates do you take the reciprocal of that combined rate to get the time, provided that rate is positive. The argument extends without any change to three or more helpers, and even to helpers who work against each other. A drain that empties a tank is just a negative rate, and it enters the same sum with a minus sign.

Working together

To find how long a team takes, add the individual rates to get the combined rate, then take the reciprocal of that sum. The adding step is exactly the algebraic-fraction addition from the first chapter: rewrite the unit fractions over a common denominator and combine.

Worked example 1 Two painters, one room

One painter can paint a room alone in $4$ hours; a second painter can paint the same room alone in $6$ hours. Working together, how long do they take?

Turn each time into a rate. The first paints $\frac{1}{4}$ of the room per hour, the second paints $\frac{1}{6}$ per hour. Add the rates over the common denominator $12$ to get the combined rate:

\frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \text{ of the room per hour}.

Together they paint $\frac{5}{12}$ of the room each hour. The time for the whole room is the reciprocal of that rate:

T = \frac{1}{\,5/12\,} = \frac{12}{5} = 2.4 \text{ hours}.

So they finish in $\frac{12}{5}$ hours, which is $2$ hours and $24$ minutes (since $0.4 \times 60 = 24$ ). Notice the answer is smaller than either painter’s solo time of $4$ or $6$ hours, exactly as it must be: two people working together beat either one alone.

In one hour, painter A finishes 1/4 of the job (3 of the 12 equal parts) and painter B finishes 1/6 (2 parts). Their rates add to 5/12 of the job per hour, so the team does more each hour than either painter alone.

Check your understanding

One hose fills a pool in $3$ hours; another fills the same pool in $6$ hours. Running together, how long do they take to fill it?

4.5

hours B

9

hours C

2

hours D

3

hours

Finding an unknown time

The same equation runs in reverse. If you know the team’s time and every individual time but one, the missing rate is whatever remains after you subtract the known rates from the combined rate. Solve for that one rate, then flip it to get the time. Because the unknown enters as a rate, the equation stays linear even though it is written with fractions.

Worked example 2 How long would the second worker take?

Two people raking leaves finish a yard together in $6$ hours. Working alone, the first would take $10$ hours. How long would the second take working alone?

Let the second worker’s solo time be $T_2$ hours, so the second rate is $\frac{1}{T_2}$ . The combined rate is $\frac{1}{6}$ per hour and the first rate is $\frac{1}{10}$ per hour, and the rates add:

\frac{1}{10} + \frac{1}{T_2} = \frac{1}{6}.

Solve for the unknown rate by subtracting the known one, using the common denominator $30$ :

\frac{1}{T_2} = \frac{1}{6} - \frac{1}{10} = \frac{5}{30} - \frac{3}{30} = \frac{2}{30} = \frac{1}{15}.

The second worker’s rate is $\frac{1}{15}$ of the yard per hour, so the solo time is the reciprocal:

T_2 = 15 \text{ hours}.

Check it: $\frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6}$ , the combined rate, so the pair really does finish in $6$ hours.

Filling against a drain

A rate can be negative. A pipe that fills a tank adds a positive rate; a drain that empties it contributes a negative one, because it undoes work rather than doing it. The net rate is still just the signed sum of the separate rates. The time to fill the tank is the reciprocal of that net rate, provided the net rate is positive. If the drain were the faster of the two, the net rate would be negative and the tank would never fill at all.

Worked example 3 A pipe and an open drain

A pipe fills an empty tank in $4$ hours. A drain at the bottom, left open, would empty a full tank in $6$ hours. If the pipe runs while the drain stays open, how long does the tank take to fill?

The pipe works at $+\frac{1}{4}$ tank per hour. The drain removes water, so it works at $-\frac{1}{6}$ tank per hour. Add the signed rates over the common denominator $12$ to get the net rate:

\frac{1}{4} - \frac{1}{6} = \frac{3}{12} - \frac{2}{12} = \frac{1}{12} \text{ tank per hour}.

The net rate is positive, so the tank does fill, gaining $\frac{1}{12}$ of a tank each hour. The time is the reciprocal of the net rate:

T = \frac{1}{\,1/12\,} = 12 \text{ hours}.

So the tank fills in $12$ hours. The open drain stretches a $4$ -hour fill out to $12$ hours, which fits the picture: some of every hour’s inflow is lost back out the bottom.

Check your understanding

A faucet fills a basin in $3$ hours. With the plug out, a drain would empty the full basin in $6$ hours. With the faucet on and the drain open, how long does the basin take to fill?

1.5

hours B

6

hours C

2

hours D

9

hours

Distance, rate, and time

The second family of rate problems trades “jobs per hour” for “miles per hour,” but the structure is identical. Distance covered is rate multiplied by time:

d = rt.

Here $r$ is the speed, the distance covered per unit of time, and it plays the same role the work rate did. One relationship gives you all three questions, because you can solve it for whichever letter is unknown:

r = \frac{d}{t}, \qquad t = \frac{d}{r}, \qquad d = rt.

Worked example 4 A steady drive, two ways

A delivery van covers $165$ miles in $3$ hours at a steady speed. First find that speed, then find how far the van travels in $7$ hours at the same speed.

The speed is distance divided by time:

r = \frac{d}{t} = \frac{165}{3} = 55 \text{ miles per hour}.

Now hold that speed and run the relationship the other way, with $t = 7$ :

d = rt = 55 \times 7 = 385 \text{ miles}.

So the van travels at $55$ mph and covers $385$ miles in $7$ hours. The single relationship $d = rt$ answered both questions; you only changed which letter you solved for.

Average speed done right

When a trip runs at one speed for a while and a different speed for the rest, its average speed is not the average of the two speeds. Average speed has one definition, and you must always go back to it:

\text{average speed} = \frac{\text{total distance}}{\text{total time}}.

Add up all the distance, add up all the time, and divide. The reason the mean of the two speeds gives the wrong answer is worth pinning down, because the trap is so easy to fall into.

Why average speed is total distance over total time#

Average speed answers one question. If the whole trip had been run at a single steady speed, what speed would cover the same distance in the same total time? That is the speed you could replace the trip with and change nothing overall. So that speed must be the total distance divided by the total time, by the very definition $r = d/t$ applied to the trip as a whole.

Now see why the plain average of the two speeds is almost always too high. Suppose you cover equal distances at a slow speed and a fast speed. Because time is distance divided by speed, the slow leg takes more time than the fast leg: the slower you go, the longer you spend going. So when you form total distance over total time, the total time is weighted toward the slow leg, where you lingered. That weighting drags the average down toward the slower speed. The plain mean of the two speeds pretends you spent equal time at each, but you did not. You spent equal distance, and more time at the slower speed. That extra time at the low speed is exactly what pulls the true average below the halfway mark. Two cases, and no others, land the average exactly on the simple mean. They are when the two speeds are equal, and when you spend equal time (not equal distance) at each leg.

Worked example 5 There and back at different speeds

You drive to a town $60$ miles away at $30$ mph, then return home along the same road at $60$ mph. What is your average speed for the whole round trip?

Resist averaging $30$ and $60$ to get $45$ . Go back to the definition, and build the total distance and the total time from the two legs.

The total distance is the round trip:

d = 60 + 60 = 120 \text{ miles}.

Now the time for each leg, using $t = d/r$ . The trip out at $30$ mph takes $\frac{60}{30} = 2$ hours; the trip back at $60$ mph takes $\frac{60}{60} = 1$ hour. So the total time is

t = 2 + 1 = 3 \text{ hours}.

Divide total distance by total time:

\text{average speed} = \frac{120}{3} = 40 \text{ miles per hour}.

The average is $40$ mph, below the naive $45$ mph. That is because you spent $2$ of the $3$ hours crawling at the slower $30$ mph and only $1$ hour at $60$ mph. More time at the low speed pulls the average down.

Check your understanding

A cyclist rides $12$ miles out at $12$ mph and returns the same $12$ miles at $4$ mph. What is the average speed for the whole ride?

8

mph B

6

mph C

5

mph D

9

mph

Common mistakes

Averaging or adding the times. If one worker takes $4$ hours and another $6$ , the team does not take the average $5$ hours, nor the sum $10$ hours. Add the rates $\frac{1}{4} + \frac{1}{6}$ , then take the reciprocal. The together time is always shorter than the faster worker alone.
Forgetting to flip the rate back to a time. After adding rates you have a combined rate, like $\frac{5}{12}$ of the job per hour. The question asks for a time, so you still must take the reciprocal: $\frac{12}{5}$ hours, not $\frac{5}{12}$ .
Adding a drain instead of subtracting it. A drain undoes work, so it enters the sum as a negative rate. Filling at $\frac{1}{4}$ against a drain of $\frac{1}{6}$ gives a net $\frac{1}{4} - \frac{1}{6}$ , not $\frac{1}{4} + \frac{1}{6}$ .
Averaging the two speeds. For a trip at two speeds, average speed is total distance over total time, never the mean of the speeds. Going $30$ mph then $60$ mph over equal distances averages $40$ mph, not $45$ , because more time is spent at the slower speed.
Mixing up the parts of $d = rt$ . Keep the units straight: a speed is miles per hour, a time is hours, and their product is miles. If you are given a distance and a speed, you divide to get a time; do not multiply everything in sight.

Key takeaways

A job finished in time $T$ is done at the rate $\frac{1}{T}$ per unit time, and rate and time are reciprocals: $\text{time} = \frac{1}{\text{rate}}$ .
Rates add when workers combine: $\frac{1}{T_1} + \frac{1}{T_2} = \frac{1}{T}$ . Add the rates, then take the reciprocal to get the together time $T$ .
To find a missing worker’s time, subtract the known rates from the combined rate, then flip the leftover rate. Because the unknown is a rate, the equation stays linear.
A drain is a negative rate: the net rate is the signed sum, and the fill time is its reciprocal when the sum is positive.
Distance, rate, time: $d = rt$ , so $r = \frac{d}{t}$ and $t = \frac{d}{r}$ ; solve for whichever one is unknown.
Average speed is $\dfrac{\text{total distance}}{\text{total time}}$ , not the mean of the speeds. Over equal distances at two speeds it falls below the mean, because more time is spent at the slower speed.

Rate and Work Problems

Learning goals

From time to rate

Why work rates add

Working together

Finding an unknown time

Filling against a drain

Distance, rate, and time

Average speed done right

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)